Question

2.
Se consideră suma S =
$\frac{1}{\sqrt{2} + 1 } + \frac{1}{ \sqrt{3} + \sqrt{2} } + ... + \frac{1}{ \sqrt{x + 1} + \sqrt{x} }$

a) Să se calculeze suma S.
b) Să se determine n apartine N, pentru care S=2018.
Va roooog!Dau Coroana!!​

Answer 1

Răspuns:

Explicație pas cu pas:

$S=...=\dfrac{\sqrt{2}-1 }{(\sqrt{2}+1)(\sqrt{2}-1)} +\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2} )(\sqrt{3}-\sqrt{2})} +...+\dfrac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}+\sqrt{x} )(\sqrt{x+1}-\sqrt{x})}=\\=\dfrac{\sqrt{2}-1 }{(\sqrt{2})^2-1^2} +\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} +...+\dfrac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1})^2-(\sqrt{x})^2}=$$=\dfrac{\sqrt{2}-1 }{2-1} +\dfrac{\sqrt{3}-\sqrt{2}}{3-2} +...+\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{x+1}-\sqrt{x}=\sqrt{x+1}-1$

Deci, S=√(x+1) - 1.

b) S=2018, ⇒ √(x+1) - 1=2018, ⇒ √(x+1) = 2018+1, ⇒√(x+1) = 2019 |², ⇒ x+1=2019², ⇒ x=2019²-1=4076361 - 1 = 4076360.